Q42E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

A differential equation and a nontrivial solution f are given. Find a second linearly independent solution using reduction of order.

${\mathbf{t}}^{\mathbf{2}}\mathbf{y}\text{'}\text{'}\mathbf{+}\mathbf{6}\mathbf{ty}\mathbf{\text{'}}\mathbf{+}\mathbf{6}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{,} \mathbf{t}\mathbf{>}\mathbf{0}\mathbf{;} \mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{t}}^{\mathbf{-}\mathbf{2}}$

Step-by-Step Solution

Answer

The second linearly independent solution of the given equation

${\mathrm{t}}^2\mathrm{y}\text{'}\text{'}+6\mathrm{ty}\text{'}+6\mathrm{y}=0, \mathrm{t}>0; \mathrm{f}\left(\mathrm{t}\right)={\mathrm{t}}^{-2}$ is $\mathrm{y}={\mathrm{c}}_1{\mathrm{t}}^{-2}+{\mathrm{c}}_2{\mathrm{t}}^{-3}$.

Step 1: Substitute the values

Given differential equation is ${\mathrm{t}}^2\mathrm{y}\text{'}\text{'}+6\mathrm{ty}\text{'}+6\mathrm{y}=0$ then the standard form of the solution is $\mathrm{y}\text{'}\text{'}+\frac{6}{\mathrm{t}}\mathrm{y}\text{'}+\frac{6}{{\mathrm{t}}^2}\mathrm{y}=0$ and $\mathrm{f}\left(\mathrm{t}\right)={\mathrm{t}}^{-2}$ is one of the solutions and $\mathrm{p}\left(\mathrm{t}\right)=\frac{6}{\mathrm{t}}$. Then

$\begin{array}{rcl}{\mathrm{y}}_2& =& {\mathrm{y}}_1\left(\mathrm{t}\right)\int \frac{{\mathrm{e}}^{-\int \mathrm{p}\left(\mathrm{t}\right)}\mathrm{dt}}{{\mathrm{y}}_1^2\left(\mathrm{t}\right)}\mathrm{dt}\\ & =& {\mathrm{t}}^{-2}\int \frac{{\mathrm{e}}^{-\int \frac{6}{\mathrm{t}}}\mathrm{dt}}{{\left({\mathrm{t}}^{-2}\right)}^2}\mathrm{dt}\end{array}$

Step 2: Simplification

Simplify the above and get;

$$\begin{gathered} ={\mathrm{t}}^{-2}\int \frac{{\mathrm{e}}^{\ln \left({\mathrm{t}}^{-6}\right)}}{{\mathrm{t}}^{-4}} \\ ={\mathrm{t}}^{-2}\times \int \frac{{\mathrm{t}}^{-6}}{{\mathrm{t}}^{-4}} \\ ={\mathrm{t}}^{-2}\times \left(-{\mathrm{t}}^{-1}\right) \\ =-{\mathrm{t}}^{-3} \end{gathered}$$

So, the solution is $\mathrm{y}={\mathrm{c}}_1{\mathrm{t}}^{-2}+{\mathrm{c}}_2{\mathrm{t}}^{-3}$.